Curvature is tensorial

This article gives the statement, and possibly proof, that a map constructed in a certain way is tensorial
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Statement

Let $\mathrm{\nabla }$ be a connection on a vector bundle $E$ over a differential manifold $M$. The Riemann curvature tensor of $\mathrm{\nabla }$ is given as a map $\mathrm{\Gamma }(TM)\otimes \mathrm{\Gamma }(TM)\otimes \mathrm{\Gamma }(E)\to \mathrm{\Gamma }(E)$ defined by:

$R(X,Y)Z={\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-{\mathrm{\nabla }}_{[X,Y]}Z$

We claim that $R$ is a tensorial map in each of the variables $X,Y,Z$.

Related facts

• Curvature is antisymmetric in first two variables

Facts used

• Leibniz rule for derivations: This states that for a vector field $X$ and functions $f,g$, we have:

$X(fg)=(Xf)(g)+f(Xg)$

• Corollary of Leibniz rule for Lie bracket: This states that for a function $f$ and vector fields $X,Y$:

$f[X,Y]=[fX,Y]+(Yf)X$

$f[X,Y]=[X,fY]-(Xf)Y$

• The Leibniz rule axiom that's part of the definition of a connection, namely:

${\mathrm{\nabla }}_X(fZ)=(Xf)(Z)+f{\mathrm{\nabla }}_X(Z)$

Proof

To prove tensoriality in a variable, it suffices to show $C^{\mathrm{\infty }}$-linearity in that variable. This is because linearity in $C^{\mathrm{\infty }}$-functions guarantees linearity in a function that is 1 at exactly one point, and zero at others.

The proofs for $X$ and $Y$ are analogous, and rely on manipulation of the Lie bracket $[fX,Y]$ and the property of a connection being $C^{\mathrm{\infty }}$ in the subscript vector. These proofs do not involve any explicit use of $Z$. The proof for $Z$ relies simply on repeated application of the product rule, and the fact that $XY-YX=[X,Y]$.

Tensoriality in the first variable

Let $f:M\to \mathbb{R}$ be a scalar function. We will show that:

$R(fX,Y)=fR(X,Y)$

We start out with the left side:

${\mathrm{\nabla }}_{fX}{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{fX}-{\mathrm{\nabla }}_{[fX,Y]}$

Now by the definition of a connection, $\mathrm{\nabla }$ is $C^{\mathrm{\infty }}$-linear in its subscript argument. Thus, the above expression can be written as:

$f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y(f{\mathrm{\nabla }}_X)-{\mathrm{\nabla }}_{[fX,Y]}$

Now applying the Leibniz rule for connections, we get:

$f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-(Yf){\mathrm{\nabla }}_X-f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[fX,Y]}$

We can rewrite $(Yf){\mathrm{\nabla }}_X={\mathrm{\nabla }}_{(Yf)X}$ and we then get:

$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X)-{\mathrm{\nabla }}_{(Yf)X+[fX,Y]}$

By the corollary stated above, we have:

$(Yf)X+[fX,Y]=f[X,Y]$

which, substituted back, gives:

$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[X,Y]})$

Tensoriality in the second variable

Let $f:M\to \mathbb{R}$ be a scalar function. We will show that:

$R(X,fY)=fR(X,Y)$

We start out with the left side:

${\mathrm{\nabla }}_X{\mathrm{\nabla }}_{fY}-{\mathrm{\nabla }}_{fY}{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[X,fY]}$

Applying the Leibniz rule and the property of a connection being $C^{\mathrm{\infty }}$ in its subscript variable yields:

$(Xf){\mathrm{\nabla }}_Y+f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X)-{\mathrm{\nabla }}_{[X,fY]}$

which simplifies to:

$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X)-{\mathrm{\nabla }}_{[X,fY]-(Xf)Y}$

We now use the corollary stated above:

$f[X,Y]=[X,fY]-(Xf)Y$

substituting this gives:

$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[X,Y]}$

which is $fR(X,Y)$

Tensoriality in the third variable

Let $f:M\to \mathbb{R}$ be a scalar function. We will show that:

$R(X,Y)(fZ)=fR(X,Y)Z$

We start out with the left side:

${\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y(fZ)-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X(fZ)-{\mathrm{\nabla }}_{[X,Y]}(fZ)$

Now we apply the Leibniz rule for connnections on each term:

${\mathrm{\nabla }}_X((Yf)(Z)+f{\mathrm{\nabla }}_{Y}Z)-{\mathrm{\nabla }}_Y((Xf)Z+f{\mathrm{\nabla }}_{X}Z)-f{\mathrm{\nabla }}_{[X,Y]}Z-([X,Y]f)Z$

We again apply the Leibniz rule to the first two term groups:

$(XYf)(Z)+(Yf){\mathrm{\nabla }}_{X}Z+(Xf){\mathrm{\nabla }}_{Y}Z+f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-(YXf)Z-(Xf){\mathrm{\nabla }}_{Y}Z-(Yf){\mathrm{\nabla }}_{X}Z-f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-f{\mathrm{\nabla }}_{[X,Y]}Z-([X,Y]f)Z$

After cancellations we are left with the following six terms:

$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[X,Y]})Z+((XY-YX-[X,Y])f)Z$

But since $[X,Y]=XY-YX$, the last three terms vanish, and we are left with:

$fR(X,Y)Z$